\(\int \frac {x^3}{(a+b x^4)^{5/4}} \, dx\) [1143]

Optimal result

Integrand size = 15, antiderivative size = 16 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{b \sqrt [4]{a+b x^4}} \]

[Out]

-1/b/(b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{b \sqrt [4]{a+b x^4}} \]

[In]

Int[x^3/(a + b*x^4)^(5/4),x]

[Out]

-(1/(b*(a + b*x^4)^(1/4)))

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{b \sqrt [4]{a+b x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{b \sqrt [4]{a+b x^4}} \]

[In]

Integrate[x^3/(a + b*x^4)^(5/4),x]

[Out]

-(1/(b*(a + b*x^4)^(1/4)))

Maple [A] (verified)

Time = 4.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
gosper	\(-\frac {1}{b \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(15\)
derivativedivides	\(-\frac {1}{b \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(15\)
default	\(-\frac {1}{b \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(15\)
trager	\(-\frac {1}{b \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(15\)
pseudoelliptic	\(-\frac {1}{b \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(15\)

[In]

int(x^3/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-1/b/(b*x^4+a)^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{b^{2} x^{4} + a b} \]

[In]

integrate(x^3/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

-(b*x^4 + a)^(3/4)/(b^2*x^4 + a*b)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=\begin {cases} - \frac {1}{b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3/(b*x**4+a)**(5/4),x)

[Out]

Piecewise((-1/(b*(a + b*x**4)**(1/4)), Ne(b, 0)), (x**4/(4*a**(5/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b} \]

[In]

integrate(x^3/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

-1/((b*x^4 + a)^(1/4)*b)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b} \]

[In]

integrate(x^3/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

-1/((b*x^4 + a)^(1/4)*b)

Mupad [B] (verification not implemented)

Time = 6.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{b\,{\left (b\,x^4+a\right )}^{1/4}} \]

[In]

int(x^3/(a + b*x^4)^(5/4),x)

[Out]

-1/(b*(a + b*x^4)^(1/4))